php - 代码=3840 "Invalid value around character 0."

Question

我正在尝试在 iOS 中创建一个应用程序，但我在注册用户时遇到问题，当我点击注册按钮时，我收到代码 3840 错误，“字符 0 周围的值无效”。我正在使用 Swift、PHP 和 JSON

这是我的代码

RegistrarUsuario.swift

@IBAction func RegistrarUsuario(sender: AnyObject) {

    let usuarioCorreo = Correo.text
    let usuarioNombre = Nombre.text
    let usuarioApellido = Apellido.text
    let usuarioContrasena = Contrasena.text


    if (usuarioCorreo!.isEmpty || usuarioApellido!.isEmpty || usuarioNombre!.isEmpty || usuarioContrasena!.isEmpty)
    {
        mostrarMensaje("No ha llenado todos los campos")
        return
    }

    let urlServer = NSURL(string: "http://...../..../registro.php")

   // let post:NSString = "correo=\(usuarioCorreo)&nombre=\(usuarioNombre)&apellido=\(usuarioApellido)&contraseña=\(usuarioContrasena)"

    let post:NSDictionary = ["correo": usuarioCorreo!, "nombre": usuarioNombre!, "apellido": usuarioApellido!, "contrasena": usuarioContrasena!]
    NSLog("PostData: %@",post);

    //let postData:NSData = post.dataUsingEncoding(NSASCIIStringEncoding)!
    let postData:NSData = NSKeyedArchiver.archivedDataWithRootObject(post)
    let postLength:NSString = String( postData.length )
    let request:NSMutableURLRequest = NSMutableURLRequest(URL: urlServer!)
    request.HTTPMethod = "POST"
    request.HTTPBody = postData
    request.setValue(postLength as String, forHTTPHeaderField: "Content-Length")
    request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
    request.setValue("application/json; charset=utf-8", forHTTPHeaderField: "Accept")

    var reponseError: NSError?
    var response: NSURLResponse?

    var urlData: NSData?
    do {
        urlData = try NSURLConnection.sendSynchronousRequest(request, returningResponse:&response)
    } catch let error as NSError {
        reponseError = error
        urlData = nil
    }

    if ( urlData != nil ) {
        let res = response as! NSHTTPURLResponse!;


        NSLog("Codigo de respuesta: %ld", res.statusCode);

        if (res.statusCode >= 200 && res.statusCode < 300)
        {
            //let responseData:NSString  = NSString(data:urlData!, encoding:NSUTF8StringEncoding)!

            //NSLog("Respuesta ==> %@", responseData);

            var error: NSError?

            let jsonData:NSDictionary = try! NSJSONSerialization.JSONObjectWithData(urlData!, options:NSJSONReadingOptions.AllowFragments) as! NSDictionary


            let success:NSInteger = jsonData.valueForKey("success") as! NSInteger


            NSLog("Success: %ld", success);

            if(success == 1)
            {
                NSLog("Se ha creado la cuenta correctamente!");
                self.dismissViewControllerAnimated(true, completion: nil)
            } else {
                var error_msg:NSString

                if jsonData["error_message"] as? NSString != nil {
                    error_msg = jsonData["error_message"] as! NSString
                } else {
                    error_msg = "Error desconocido"
                }
                let alertView:UIAlertView = UIAlertView()
                alertView.title = "No se pudo crear la cuenta!"
                alertView.message = error_msg as String
                alertView.delegate = self
                alertView.addButtonWithTitle("Aceptar")
                alertView.show()

            }

        } else {
            let alertView:UIAlertView = UIAlertView()
            alertView.title = "No se pudo crear la cuenta!"
            alertView.message = "Conexion Fallida"
            alertView.delegate = self
            alertView.addButtonWithTitle("Aceptar")
            alertView.show()
        }
    }  else {
        let alertView:UIAlertView = UIAlertView()
        alertView.title = "No se pudo crear la cuenta!"
        alertView.message = "Conexion Fallida"
        if let error = reponseError {
            alertView.message = (error.localizedDescription)
        }
        alertView.delegate = self
        alertView.addButtonWithTitle("Aceptar")
        alertView.show()
    }


}

Registro.php

<?php

header('Content-type: application/json');

if($_POST) {

    $correo  = $_POST['correo'];
    $nombre   = $_POST['nombre'];
    $apellido = $_POST['apellido'];
    $contraseña = $_POST['contraseña'];
    if($_POST['correo']) {


        $mysqli = new mysqli('localhost', 'root', '', 'prueba');

        /* check connection */
        if (mysqli_connect_errno()) {
            error_log("Conexion fallida: " . mysqli_connect_error());
            echo '{"success":0,"error_message":"' . mysqli_connect_error() . '"}';
        } else {
            $query = "INSERT INTO usuarios (correo, nombre, apellido, contrasena) VALUES (?, ?, ?, ?)";
            if($stmt = $mysqli->prepare($query)){
            $contrasena = md5($contrasena);
            $stmt->bind_param('ss', $correo, $nombre, $apellido, $contrasena);

            /* execute prepared statement */
            $stmt->execute();
            }
            else
            {
                die("Errormessage: ". $mysqli->error);
            }
            if ($stmt->error) {error_log("Error: " . $stmt->error); }

            $success = $stmt->affected_rows;

            /* close statement and connection */
            $stmt->close();

            /* close connection */
            $mysqli->close();
            error_log("Success: $success");

            if ($success > 0) {
                error_log("User '$correo' ha sido creado.");
                echo '{"success":1}';
            } else {
                echo '{"success":0,"error_message":"Correo Existe."}';
            }
        }
} else {
    echo '{"success":0,"error_message":"Usuario Invalido."}';
}
}else {
    echo '{"success":0,"error_message":"Informacion Invalida."}';
}
?>

非常感谢您的帮助，谢谢。

Answer 1

最近我也遇到了这个问题。

只需再次查看您的响应/当您执行您正在为其执行响应的 Web 服务时。

在您的情况下，只需尝试“responseString 而不是responseJSON”。

希望对你有帮助。